Midterm 2

# Midterm 2 - Version 028/AABDA – midterm 02 – Turner...

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Unformatted text preview: Version 028/AABDA – midterm 02 – Turner – (59130) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points After a 5 . 65 Ω resistor is connected across a battery with a 0 . 08 Ω internal resistance, the electric potential between the physical battery terminals is 7 V. What is the rated emf of the battery? 1. 13.5033 2. 10.4897 3. 10.1571 4. 7.09912 5. 12.9461 6. 13.175 7. 9.43299 8. 11.5814 9. 15.4682 10. 7.45337 Correct answer: 7 . 09912 V. Explanation: Let : R = 5 . 65 Ω , r = 0 . 08 Ω , and V = 7 V . The current drawn by the external resistor is given by I = V R = 7 V 5 . 65 Ω = 1 . 23894 A . The output voltage is reduced by the inter- nal resistance of the battery by V = E − I r , so the electromotive force is E = V + I r = 7 V + (1 . 23894 A) (0 . 08 Ω) = 7 . 09912 V . 002 10.0 points A capacitor network with air-filled capacitors as shown below. 80 . 7 V 25 . 8 μ F 25 . 8 μ F 25 . 8 μ F 25 . 8 μ F b a c d When the top right-hand capacitor is filled with a material of dielectric constant κ , the charge on this capacitor is increases by a fac- tor of 1 . 24. Find the dielectric constant κ of the mate- rial inserted into the top right-hand capacitor. 1. 1.5641 2. 1.5 3. 1.98507 4. 2.1746 5. 2.50877 6. 2.27869 7. 2.0303 8. 1.53165 9. 3.0 10. 1.63158 Correct answer: 1 . 63158. Explanation: Let : C 1 = C = 25 . 8 μ F , C 2 = C = 25 . 8 μ F , C 3 = C = 25 . 8 μ F , C 4 = C = 25 . 8 μ F , E B = 80 . 7 V , and Q ′ = 1 . 5 Q . Version 028/AABDA – midterm 02 – Turner – (59130) 2 E B C 1 C 3 C 2 C 4 b a c d The capacitors C 3 and C 4 have nothing to do with this problems. In addition, the capac- itances are all equal and their specific values are immaterial. Furthermore, the electric po- tential of the battery is not required. C 1 = C 2 = C 3 = C 4 , where Q and Q ′ are the initial and final charges on C 2 and Q ′ Q ≡ α =ratio of final to initial charge on C 2 . We know the charges on C 1 and C 2 are the same. Initially, V ab = V 1 + V 2 = Q C 1 + Q C 2 = Q C + Q C = 2 Q C . (1) Therefore Q = 1 2 V ab C . After the dielectric material is inserted in C 2 , the capacitance becomes C ′ 2 = κC . There- fore, V ab = V ′ 1 + V ′ 2 = Q ′ C 1 + Q ′ C ′ 2 = Q ′ C + Q ′ κC = κ + 1 κ Q ′ C , and using Eq. (1) and solving for Q ′ , we have 2 Q C = κ + 1 κ Q ′ C Q ′ = κ κ + 1 V ab C = κ κ + 1 2 Q Q ′ Q ≡ α = 2 κ κ + 1 = 1 . 24 . Solving for κ , we have κ = α 2 − α = 1 . 24 2 − 1 . 24 = 1 . 63158 . keywords: 003 10.0 points An air-filled cylindrical capacitor has a capac- itance of 16 pF and is 5 . 9 cm in length....
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Midterm 2 - Version 028/AABDA – midterm 02 – Turner...

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